Question

The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is: (taking \[h = 6 \times {10^{ - 34}}Js\])
A. 100
B. \[3 \times {10^{20}}\]
C. \[3 \times {10^{18}}\]
D. \[1000\]

Answer 1

Hint:The photon is the qualitative unit of energy of the light wave. It is proportional to the frequency of the light wave. The frequency of the light wave is the most characteristic property because it doesn’t change by changing the medium in which the light is travelling.

Formula used:
\[E = h\nu \]
where h is the Plank’s constant and E is the energy of the photon with frequency equal to \[\nu \].
\[c = \nu \lambda \]
where c is the speed of light, \[\nu \]is the frequency of the photon and \[\lambda \]is the wavelength of the light wave.

Complete step by step solution:
The wavelength of the photon is given as 540 nm.
As we know that \[1\,nm = {10^{ - 9}}\,m\]
The wavelength of the given photon in S.I. unit will be,
\[\lambda = 540 \times {10^{ - 9}}m\]
\[\Rightarrow \lambda = 5.40 \times {10^{ - 7}}m\]
We got the wavelength of the photon and now we need to use the relation of the characteristic physical quantities for the electromagnetic wave to find the frequency of the photon,
\[c = \nu \lambda \]
\[\Rightarrow \nu = \dfrac{c}{\lambda }\]

On putting the value of the speed of light and the wavelength of the photon, we get
\[\nu = \dfrac{{3 \times {{10}^8}}}{{5.40 \times {{10}^{ - 7}}}}Hz\]
\[\Rightarrow \nu = 5.55 \times {10^{14}}Hz\]
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
Putting the values of Plank’s constant and the frequency, we get
\[E = \left( {6 \times {{10}^{ - 34}}} \right) \times \left( {5.55 \times {{10}^{14}}} \right)J\]
\[\Rightarrow E = 3.33 \times {10^{ - 19}}J\]
Hence, there is \[3.33 \times {10^{ - 19}}J\] energy per second by each photon.

If there are n photons per second, then total energy per second will be,
\[E = 3.33n \times {10^{ - 19}}J\]
As we know that the energy per unit time is equal to the power of the source.
\[3.33n \times {10^{ - 19}} = 100\]
\[\Rightarrow n = \dfrac{{100}}{{3.33 \times {{10}^{ - 19}}}}\]
\[\therefore n = 3 \times {10^{18}}\]
Hence, the emitted number of photons per second is \[3 \times {10^{18}}\].

Therefore, the correct option is C.

Note: We must be careful about the units of the physical quantity while solving the numerical problem. We need to convert all the given data into the standard unit form. The photon's energy is expressed either in Joule as per S.I unit or in electron-Volt. One electron-Volt is the energy possessed by an electron when it is accelerated through the potential difference of unit volt.